When 100 V d.c. is applied across a solenoid, a current of 1 A flows in it. When 100 a.c. is applied across it, the current drops to 0.5 A . If the frequency is 50 Hz , the impedance and inductance is

Step-by-step solution

When dc is applied i= V R 1= 100 R R=100 When ac is applied i= V Z 0.5= 100 Z Z=200 Hence Z= R^2+X_L^2 = R^2+4 ^2 v^2 L^2 aligned & (200)^2=(100)^2+4 ^2(50)^2 L^2 & L=0.55 H aligned